Bài 4.19 trang 204 SBT giải tích 12

Thực hiện các phép tính sau:

a) \(\dfrac{{(2 + i) + (1 + i)(4 - 3i)}}{{3 + 2i}}\)

b) \(\dfrac{{(3 - 4i)(1 + 2i)}}{{1 - 2i}} + 4 - 3i\)

Lời giải

a) \(\dfrac{{(2 + i) + (1 + i)(4 - 3i)}}{{3 + 2i}}\)\( = \dfrac{{2 + i + 7 + i}}{{3 + 2i}} = \dfrac{{9 + 2i}}{{3 + 2i}}\) \( = \dfrac{{\left( {9 + 2i} \right)\left( {3 - 2i} \right)}}{{\left( {3 + 2i} \right)\left( {3 - 2i} \right)}} = \dfrac{{31 - 12i}}{{13}}\) \( = \dfrac{{31}}{{13}} - \dfrac{{12}}{{13}}i\)

b) \(\dfrac{{(3 - 4i)(1 + 2i)}}{{1 - 2i}} + 4 - 3i\)\( = \dfrac{{11 + 2i}}{{1 - 2i}} + 4 - 3i\)\( = \dfrac{{\left( {11 + 2i} \right)\left( {1 + 2i} \right)}}{{\left( {1 - 2i} \right)\left( {1 + 2i} \right)}} + 4 - 3i\)  \( = \dfrac{{7 + 24i}}{{1 + 4}} + 4 - 3i\) \( = \dfrac{7}{5} + \dfrac{{24}}{5}i + 4 - 3i\) \( = \dfrac{{27}}{5} + \dfrac{9}{5}i\)